Calculus Examples

$\frac{d y}{d x} = \frac{1}{x}$ and $y (e) = - 2$

Step 1

Rewrite the equation.

$d y = \frac{1}{x} d x$

Step 2

Integrate both sides.

Tap for more steps...

Step 2.1

Set up an integral on each side.

$\int d y = \int \frac{1}{x} d x$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int \frac{1}{x} d x$

Step 2.3

The integral of $\frac{1}{x}$ with respect to $x$ is $\ln (| x |)$ .

$y + C_{1} = \ln (| x |) + C_{2}$

Step 2.4

Group the constant of integration on the right side as $C$ .

$y = \ln (| x |) + C$

Step 3

Use the initial condition to find the value of $C$ by substituting $e$ for $x$ and $- 2$ for $y$ in $y = \ln (| x |) + C$ .

$- 2 = \ln (| e |) + C$

Step 4

Solve for $C$ .

Tap for more steps...

Step 4.1

Rewrite the equation as $\ln (| e |) + C = - 2$ .

$\ln (| e |) + C = - 2$

Step 4.2

Simplify each term.

Tap for more steps...

Step 4.2.1

$e$ is approximately $2.71828182$ which is positive so remove the absolute value

$\ln (e) + C = - 2$

Step 4.2.2

The natural logarithm of $e$ is $1$ .

$1 + C = - 2$

Step 4.3

Move all terms not containing $C$ to the right side of the equation.

Tap for more steps...

Step 4.3.1

Subtract $1$ from both sides of the equation.

$C = - 2 - 1$

Step 4.3.2

Subtract $1$ from $- 2$ .

$C = - 3$

Step 5

Substitute $- 3$ for $C$ in $y = \ln (| x |) + C$ and simplify.

Tap for more steps...

Step 5.1

Substitute $- 3$ for $C$ .

$y = \ln (| x |) - 3$